alfo alſo angle ABC angle BAC bafe baſe BC is equal BC is given becauſe the angle becauſe the ratio bifected Book XI cafe circle ABCD circumference cone conſtruction cylinder defcribed demonſtrated drawn EFGH equal angles equiangular equimultiples Euclid excefs faid fame multiple fame ratio fecond fegment fhall fhewn fides fides BA fimilar fince firft firſt folid angle fore fquare fquare of AC given angle given in fpecies given in magnitude given in pofition given magnitude given ratio given ſtraight line gnomon greater join lefs leſs likewife line BC muſt oppofite parallel parallelepipeds parallelogram perpendicular plane angles PROP Propofition pyramid rectangle contained rectilineal figure right angles ſame ſhall ſpace ſphere ſquare ſtraight line AB THEOR theſe thro tiple triangle ABC wherefore
Seite 157 - D ; wherefore the remaining angle at C is equal to the remaining angle at F ; Therefore the triangle ABC is equiangular to the triangle DEF. Next, let each of the angles at C, F be not less than a right angle ; the triangle ABC is also, in this case, equiangular to the triangle DEF.
Seite 45 - If a straight line be divided into any two parts, the rectangle contained by the whole and one of the parts, is equal to the rectangle contained by the two parts, together with the square of the aforesaid part.
Seite 73 - THE straight line drawn at right angles to the diameter of a circle, from the extremity of...
Seite 163 - If two triangles have one angle of the one equal to one angle of the other and the sides about these equal angles proportional, the triangles are similar.
Seite 81 - ELF. Join BC, EF ; and because the circles ABC, DEF are equal, the straight lines drawn from their centres are equal: therefore the two sides BG, GC, are equal to the two EH, HF ; and the angle at G is equal to the angle at H ; therefore the base BC is equal (4.
Seite 323 - Equiangular parallelograms have to one another the ratio which is compounded of the ratios of their sides.
Seite 80 - EA : and because AD is equal to DC, and DE common to the triangles ADE, CDE, the two sides AD, DE are equal to the two CD, DE, each to each ; and the angle ADE is equal to the angle CDE, for each of them is a right angle ; therefore the base AE is equal (4.